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Unformatted text preview: Rules for Determining Order of Gene Function/Epistasis 1. Must compare 2 single mutants with a double mutant. 2. Each single mutant must have opposite phenotypes. 3. Where double mutant has phenotype of a single mutant, the corresponding gene acts downstream of the other gene. 4. You also need to know dominance and recessiveness. 5. When designing the pathway, start with downstream gene 1st!! Example: Genotype A* B* A*B* Phenotype And lets imagine... (recessive) (dominant) + + B is downstream of A, or epistatic to A. Signal A+ B+ response Now figure out bars and arrows: lets imagine.. Start with B, then do A........ Notes on Practice Problems for Genetics 320 Class: I am giving you 12 problems from past Exams and problem sets. If you understand these, you will do very well. NOTE: Not all of these problems are relevant. I will tell you here which ones are most relevant. You will learn much from the other ones as well, but some of the material we may not have covered in this class. I think we will cover yeast sufficiently. MOST RELEVANT: 5B, 5C 6B 7 (8A,B- I think you should know this at this point, but we have not talked about it explicitly.) 8C,8D 9 10B,C,D,E 11 12 (a hard problem) The exam will have some combination of these: a Lac problem, an F problem a Cancer Pathway problem T/F short answer problem Maybe a yeast epistasis problem Bruce's Problem (1 page) #1Q #1Q&A #2Q #2 Q&A #3Q #3Q&A #4 Q & A Answers #5Q #5Q&A One can identify recessive mutants directly by assaying haploids Ste#40 is a dominant mutation (tested at the restrictive condition, of course Or, for those of you who are really into tricks, ste#40 could have been linked to MATa..but I wouldn't do that to you unless I set the problem up more transparently) 1/4th Pheromone Ste42+ Ste41+ mating Ste#41 is recessive, so this diploid is mutant only for ste#42 function...must have a double mutant cell to perform epistasis tests!!! #6Q #6Q&A Name: 12. Lac Operon (30 points) #7Q page 5 A. Given the genotypes, fill in the appropriate phenotypes for LacZ and LacY. The inducer is lactose or IPTG. Genotype 1 I+O+Z+Y+ 2 IS O+ Z+ Y3 IS O+ Z- Y+ / F'I-OcZ+Y4 I+OcZ+Y-/ F' I-O+Z+Y5 I-OCZ+Y+ / F' IS O+Z-Y- Phenotypes LacZ expression LacY expression no inducer inducer no inducer inducer - + - + 10 pt, 1/2 ea B. Name one strain that indicates that OC is epistatic to IS, and what data tells you that? 2pt C. A new gene is identified that regulates the Lac operon, called Gene T. An allele Gene T-2 was analyzed. We know that Gene T+ acts in the glucose, ADC1, CRP pathway. The assays of Gene T+ function are done in a I- cell, with glucose present. (The phenotypes would be complicated by reporting LacZ expression with no glucose and with glucose column. However, you can figure this out using this data alone). i. Is Gene T-2 recessive or dominant (circle one) 2pt Strain LacZ expression + + + + + - ii. Draw the glucose, Gene T+,ADC1,CRP, LacZ, pathway, with gene order and sign of regulation (bar or arrow) 1. GeneT+ 2. GeneT-2 3. GeneT-2/ Gene T+ 4. ADC1* 5. ADC1* GeneT-2 6. ADC17 . ADC1- Gene T-2 8. CRP 9. CRP- GeneT-2 iii. Why is the combination of strains 2, 4 and 5 not informative for trying to determine the order of function of GeneT+ and ADC1+? Name: 12. Lac Operon (30 points) #7Q&A page 5 A. Given the genotypes, fill in the appropriate phenotypes for LacZ and LacY. The inducer is lactose or IPTG. Genotype 1 I+O+Z+Y+ 2 IS O+ Z+ Y3 IS O+ Z- Y+ / F'I-OcZ+Y4 I+OcZ+Y-/ F' I-O+Z+Y5 I-OCZ+Y+ / F' IS O+Z-Y- Phenotypes LacZ expression LacY expression no inducer inducer no inducer inducer + + + + + + + + + + 10 pt, 1/2 ea B. Name one strain that indicates that OC is epistatic to IS, and what data tells you that? #5, lacZ and lacY expression (credit given for either) #3 lacZ expression C. A new gene is identified that regulates the Lac operon, called Gene T. An allele Gene T-2 was analyzed. We know that Gene T+ acts in the glucose, ADC1, CRP pathway. The assays of Gene T+ function are done in a I- cell, with glucose present. (The phenotypes would be complicated by reporting LacZ expression with no glucose and with glucose column. However, you can figure this out using this data alone). 2pt i. Is Gene T-2 recessive or dominant (circle one) 2pt Strain LacZ expression + + + + + - ii. Draw the glucose, Gene T+,ADC1,CRP, LacZ, pathway, with gene order and sign of regulation (bar or arrow) Glucose iii. ADC1+ GeneT+ CRP LacZ 1. GeneT+ 2. GeneT-2 3. GeneT-2/ Gene T+ 4. ADC1* 5. ADC1* GeneT-2 6. ADC17 . ADC1- Gene T-2 8. CRP 9. CRP- GeneT-2 4pt order, 2 pt each bar/arrow=12pt overall Why is the combination of strains 2, 4 and 5 not informative for trying to determine the order of function of GeneT+ and ADC1+? Not informative because both single mutants have same phenotype. 4pt Name 15. Yeast Genetics Questions (30 points) #8Q A.Mitotic recombination is needed to generate MATa/MATa cells from a MATa/MATa cell (used for complemention tests). In the diagram below, draw how a MATa/MATa cell can be formed by mitotic recombination, starting with the G1 cell shown. In your drawing indicate the site of recombination and features of chromosome segregation if needed (10pt) Draw answer in space to right. Sly+ centromeres MATa Sly- MATa B. From the map shown, what fraction of the MATa/MATa cells will by homozygous for one of the sly alleles, and which allele will it be (2pt)? . Genotype C. Order of function. Ste91* has one mutation, Ste92* has one mutation. Ste91* and ste92* are not linked. Phenotypes of diploids and haploids are shown. i. Is ste91* dominant or recessive (circle one). ii. Is ste92* dominant or recessive (circle one) iii. Draw the pathway of the order of function, from pheromone to mating. Include bars and arrows. 1. 2. 3. 4. 5. 6. Wt Mating response (schmoo) no pheromone pheromone Ste91* Ste91*/+ - Ste92* + Ste92*/+ + Ste91* ste92* - + + + - D. Another two mutations are isolated, called ste93* and ste93**. Ste93* is dominant and fails to mate, while ste93** is recessive and always mates. State whether you can determine the mating phenotype of a MATa/MATa ste93*/ste93** cell, and if so what that phenotype would be. Name 15. Yeast Genetics Questions (30 points) #8Q&A A.Mitotic recombination is needed to generate MATa/MATa cells from a MATa/MATa cell (used for complemention tests). In the diagram below, draw how a MATa/MATa cell can be formed by mitotic recombination, starting with the G1 cell shown. In your drawing indicate the site of recombination and features of chromosome segregation if needed (10pt) Draw answer in space to right. MATa MATa 1 1 1&3 cM MATa MATa MATa Sly+ cosegregate 3 2 centromeres 3 Sly- MATa MATa MATa 4 10 pt: 4 pt G2 cell, 2 pt crossover, 2 pt 1 &3 coseg, 2pt final product (& 2 and 4, but that was not essential to show) B. From the map shown, what fraction of the MATa/MATa cells will by homozygous for one of the sly alleles, and which allele will it be (2pt)? 0- all products still heterozygous. "0" is enough. Genotype C. Order of function. Ste91* has one mutation, Ste92* has one mutation. Ste91* and ste92* are not linked. Phenotypes of diploids and haploids are shown. i. Is ste91* dominant or recessive (circle one). ii. Is ste92* dominant or recessive (circle one) iii. Draw the pathway of the order of function, from pheromone to mating. Include bars and arrows. 1. 2. 3. 4. 5. 6. Wt Mating response (schmoo) no pheromone pheromone Ste91* Ste91*/+ - Ste92* + Ste92*/+ + Ste91* ste92* - + + + - Pheromone Ste92+ Ste91+ mating 2p ea i and ii;10 pt pathway (4pt order, 6pt bars/arrows) D. Another two mutations are isolated, called ste93* and ste93**. Ste93* is dominant and fails to mate, while ste93** is recessive and always mates. State whether you can determine the mating phenotype of a MATa/MATa ste93*/ste93** cell, and if so what that phenotype would be. Yes..it fails to mate (phenotype of dominant mutant) 4 pt. #9Q #9Q&A #10Q #10Q&A #11Q #11Q&A See previous answers All act in cis and in trans If they are both recessive, do a complementation test- if a cell with both has a mutant phenotype, they are in the same gene. If one is dominant, you will have to map them and do a recombination test. + + + + + + + ADC1* for both 4 & 5 Make CAP- (CRP-) ADC1* double mutant. Phenotype should be lacZ-, because CAP (CRP) is downstream of ADC1. It would make cells phenotypically Z-. This mutation would only act in cis. Test by making an F'lac operon with CRP-binding site, Z+ and Y- with host genome with CRP+ and Z- and Y+. This cell will be phenotypically Z- and Y+. #12Q 5. "Mystery meat" question (5 points) An F'Leu1+ plasmid is present in a bacterial strain. The F' has only the Leu1 gene and has no IS nor transposon sequences. An F without IS nor transposons cannot form an Hfr. However, this F'Leu1 can form an Hfr. The F'Leu1+ is in a strain that has Arg2+ Leu1+ Ura3+ in that order. You have an F- strain that is Arg2- Leu1- Ura3-. 2pt i. ii. How does the F' integrate into the bacterial chromosome without IS nor transposon sequences? How many single crossovers does it require? Once the Hfr is formed, how would you know its an Hfr and where its integrated? 1pt iii. 2pt #12Q&A 5. "Mystery meat" question (5 points) I THINK ONE PERSON OF 261 GOT THIS ONE RIGHT THOUGH MANY GOT CLOSE. An F'Leu1+ plasmid is present in a bacterial strain. The F' has only the Leu1 gene and has no IS nor transposon sequences. An F without IS nor transposons cannot form an Hfr. However, this F'Leu1 can form an Hfr. The F'Leu1+ is in a strain that has Arg2+ Leu1+ Ura3+ in that order. You have an F- strain that is Arg2- Leu1- Ura3-. 2pt 1pt i. ii. How does the F' integrate into the bacterial chromosome without IS nor transposon sequences? Sco between Leu1 on F' and Leu1 on host chromosome (see diagram) How many single crossovers does it require? (1SCO- NO CREDIT FOR THIS ALONE) Once the Hfr is formed, how would you know its an Hfr and where its integrated? Observe high frequency transfer of either Arg2 or Ura3 into F- 2pt iii. Leu1 gene F'Leu1+ Arg2 Leu1 F Leu 1 Ura3 Sco between Leu1 genes leads to Hfr Arg2 Leu1 Ura3 Observe high frequency transfer of either Arg2+ or Ura3+. Cross to F- arg2leu1- ura3-. ...
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